Edit: This answer only deals with kingside castling; the figures have to be revised if you also consider queenside castling.
The three first steps:
There are 90 positions where n(P)=1
For White to be able to castle short at once, he needs to have Kf1, Rg1 in the initial position. Then there are 5 possible spots from the second rook, 3 for one bishop and 2 for the other one, and finally 3 remaining spots for the queen. The knights go to the last two spots.
There are 64 positions where n(P)=2
36 of them with Kf1, Ng1, Rh1: White can play 1.Nf3 or 1.Nh3 and 2.0-0
28 of them with Rg1, Nf1, Ke1: White can play 1.Ng3 or 1.Ne3 and 2.0-0
There are 236 positions where n(P)=3
please double-check this part
Here are the different configurations and the number of relevant positions:
Edit : removed 2 cases after checking the rules: the path to g1 must be open for the king (only the right-hand rook can be in the way)
Once we reach n(P)=4, we cannot ignore Black's play anymore: for instance, QNRNKBBR looks like it should have n=5, but actually it has n=4 because of the sequence 1.f4 b5 2.Bxa7 Qxg2 3.Bxg2 Nc6 4.0-0. Counting by hand becomes too fastidious and hazardous: a software program is needed.
On the other hand, after clarification by the OP, I believe 7 to be the maximum value of the function P->n(P).
If Black wasn't allowed to "help" we could build only 2 positions with n(P)=9; Here is one, the other is similar with RKBBNNQR on the first rank.
[FEN "rkqnnbbr/pppppppp/8/8/8/8/PPPPPPPP/RKQNNBBR w - - 0 1"]
In a series problem where White plays alone he would need 9 moves in order to castle. But since Black moves are taken into account, White actually can castle on move 7 from that diagram if the game goes something like 1.b3 d5 2.Qa3 Qe6 3.Nb2 Qxe2 4.Ned3 Qxf2 5.Bxf2 Nc6 6.Be2 0-0-0 7.0-0.